Bzoj 3240: [noi2013] matrix game matrix multiplication + decimal fast power + constant optimization

3240: [noi2013] matrix game time limit: 10 sec memory limit: 256 MB
Submit: 613 solved: 256
[Submit] [Status] Description

Tingting is a matrix-loving child. One day she wants to use a computer to generate a huge N-Row M-column matrix (you don't have to worry about how she stores it ). The matrix she generated satisfies a magical nature: If f [I] [J] is used to represent the elements in column J of row I in the matrix, then f [I] [J] satisfies the following recursive formula:

F [1] [1] = 1
F [I, j] = A * f [I] [J-1] + B (J! = 1)
F [I, 1] = C * f [I-1] [m] + d (I! = 1)
In the recursive formula, A, B, C, and D are given constants.

Tingting wants to know the value of F [N] [M]. Please help her. Because the final result may be very large, you only need to output the remainder of F [N] [m] divided by 1,000,000,007.

Input

A row has six integers, n, m, A, B, C, and D. Meaning as described in

Output

Contains an integer that represents the remainder of F [N] [m] divided by 1,000,000,007.

Sample input3 4 1 3 2 6 sample output85hint

 

The matrix in the example is:

1 4 7 10

26 29 32 35

76 79 82 85

 

Based on the idea of refreshing questions before noip, I made this question with no data range. I found that the range of N and M is a little too much, and the decimal power is not enough, this question should be the ordinary O (N ^ 3) matrix multiplication. Since the second row of the matrix is not changed in the question, we can optimize the constant.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MOD 1000000007#define MAXN 2100000typedef long long qword;struct matrix{        qword a[2][2];        int n,m;        matrix()        {                memset(a,0,sizeof(a));        }        void init0()        {                n=m=2;                a[0][0]=a[1][1]=1;                a[0][1]=a[1][0]=0;        }        void init1(int aa,int bb)        {                n=m=2;                a[0][0]=aa;                a[0][1]=bb;                a[1][1]=1;        }        void init2(int aa)        {                n=2;m=1;                a[0][0]=aa;                a[1][0]=1;        }};inline matrix operator *(matrix m1,matrix m2){        int i,j,k;        matrix ret;        ret.n=m1.n;        ret.m=m2.m;        if (ret.n==2 && ret.m==2)        {                if (m1.a[1][1]!=1 || m1.a[1][0]!=0 || m2.a[1][1]!=1 || m2.a[1][0]!=0)                        throw 1;                ret.a[0][0]=m1.a[0][0]*m2.a[0][0]%MOD;                ret.a[0][1]=(m1.a[0][0]*m2.a[0][1]%MOD+m1.a[0][1])%MOD;                ret.a[1][0]=0;                ret.a[1][1]=1;                return ret;        }        for (i=0;i<m1.n;i++)        {                for (j=0;j<m2.m;j++)                {                        for (k=0;k<m1.m;k++)                        {                                ret.a[i][j]=(ret.a[i][j]+m1.a[i][k]*m2.a[k][j]%MOD)%MOD;                        }                }        }        return ret;}matrix pow_mod(matrix a,char *str,int len){        int i,j;        register matrix t,l0,ret;        ret.init0();        l0.init0();        t=a;        for (i=len-1;i>=0;i--)        {                for (j=0;j<10;j++)                {                        if (j==str[i]-‘0‘)                                ret=ret*l0;                        l0=l0*t;                }                t=l0;                l0.init0();        }        return ret;}char s1[MAXN],s2[MAXN];int main(){        freopen("input.txt","r",stdin);        freopen("output.txt","w",stdout);        qword a,b,c,d,i,j,k;        scanf("%s %s%lld%lld%lld%lld",s1,s2,&a,&b,&c,&d);        a%=MOD;b%=MOD;c%=MOD;d%=MOD;        int l1,l2;        l1=strlen(s1);        l2=strlen(s2);        int x;        x=l1-1;        s1[x]--;        while (s1[x]<‘0‘)        {                s1[x]+=10;s1[x-1]--;x--;        }        x=l2-1;        s2[x]--;        while (s2[x]<‘0‘)        {                s2[x]+=10;s2[x-1]--;x--;        }        matrix m1,r1,m2,r2,r3,m3,r4,m4;        matrix t1;        m1.init1(a,b);        m2.init1(c,d);        m4.init2(1);        r1=pow_mod(m1,s2,l2);        r2=m2*r1;        r3=pow_mod(r2,s1,l1);        r4=r1*r3*m4;        cout<<r4.a[0][0]<<endl;}

 

 

 

Bzoj 3240: [noi2013] matrix game matrix multiplication + decimal fast power + constant optimization